If a given quantity of electricity liberates 0.65g of Zn2+, what amount of Hg2+ would be liberated by the same quantity of current?
[Zn = 65, Hg = 201]
- 1.00g
-
2.01g
- 4.02g
- 8.04g
The correct answer is: B
Explanation
At the cathode, we have:
\(Zn^{2+} + 2e^- → Zn_{(s)}\)
Normally, we need
2F → 65g
x → 0.65
x = 0.02F
Let us check how much mercury will be deposited by 0.02F of electricity
\(Hg^{2+} + 2e^-(aq) → Hg_{(s)}\)
2F → 201g
∴ 0.02F = x
x = 2.01g
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